Question 1041418
<pre>
This double integral is over the unit circle, from the
lower unit semicircle {{{y=-sqrt(1-x^2)}}} to the upper unit 
semicircle {{{y=sqrt(1-x^2)}}}, so we convert, using {{{x^2+y^2=r^2}}}, 
and we get:

{{{int(""^"","",0,2pi)}}}{{{int(ln(r^2+1)*r^"",dr*"d@",0,1),"",0,2pi))}}}

The radius r goes from the origin (the pole) where r is 0
out to the circumference of the unit circle, where r is 1.  
Then the angle <font face="symbol">q</font> goes around from 0 to 2<font face="symbol">p</font> . 


Use this taken from a table of integral, to save you
from having to integrate it by parts:

{{{int(ln(u),du)=u*ln(u)-u+C}}}

to complete the evaluation.  If you have trouble, tell
me in the thank-you note form below and I'll get back
to you by email.

Edwin</pre>